What is the problem with this differential equation?

Question

I am stuck a differential equation problem $x^2D^2y-2x(x+2)Dy+6(x+1)y=x^4e^x$,the question is knowing that for some $n$ positive integer ,$x^n$ is a solution of the reduced differential equation,reduce the differential equation to first order ODE.Now we can find out that $x^3$ is the solution mentioned here.So,we have to take the general solution as $y=vx^3$,then by substituting in the given differential equation,we should come up with a form $D^2v+P_1Dv=R_1$,there should be no term with $v$.But when I am substituting in the equation,I am getting $x^2D^2v+(2x-2x^2)Dv-6v=xe^x$,as per my assumption the $-6v$ term should not be there but I have checked my calculation,there is no mistake.So,what has gone wrong? Can someone tell me why I am getting stuck and where I have made mistake?

Answer 1

$$x^2y''-2x(x+2)y'+6(x+1)y=x^4e^x$$

I got this $$ \begin {cases} y=vx^3 \\ y'=v'x^3+3x^2v \\ y''=v''x^3+6x^2v'+6xv \end{cases} $$ So that we have: $$x^2y''=v''x^5+6x^4v'+\color{red}{6x^3v}$$ $$6(x+1)y=\color{blue}{6vx^4}+\color{red}{6x^3v}$$ $$-2x(x+2)y'=-2x^5v'\color{blue}{-6x^4v}-4x^4v'\color{red}{-12x^3v}$$ Now sum the terms ( the terms in v in red and blue cancel as you can see): $$v''x^5+2x^4v'-2x^5v'=x^4e^x$$ $$x^2v''+2v'x(1-x)=xe^x$$ $$(x^2v')'-2v'x^2=xe^x$$ Integrating factor is $\mu= e^{-2x}$ $$(x^2v'e^{-2x})'=xe^{-x}$$ Integrate: $$x^2v'e^{-2x}=-xe^{-x}-e^{-x}+C$$ $$v'=\dfrac 1{x^2} \left (-xe^{x}-e^{x}+Ce^{2x} \right )$$ Integrate again. You get a solution with the exponetial integral.